Can the adjugate matrix of $A$ be expressed as a polynomial of $A$?

Question

We know that if $A$ is invertible, then $A^{-1}$ can be expressed as a polynomial of $A$, that is to say, there exists a polynomial $f(x)$ such that $$ A^{-1} = f(A) $$ Of course in this case, $\operatorname{adj}(A)$ (the adjugate matrix of $A$) can also be expressed as a polynomial of $A$.

I wonder if $A$ is not invertible, can $\operatorname{adj}(A)$ be expressed as a polynomial of $A$？

Hope for your comments.

Answer 1

Yes, $\operatorname{adj}(A)$ can always be expressed as a polynomial in $A$. We can follow the same proof as for Cayley-Hamilton:

Consider $\operatorname{adj}(A - tI)$ for a scalar $t$. We have

$$(A - tI)\operatorname{adj}(A - tI) = \det(A - tI)I = p_A(t)I$$ where $p_A(t) = (-1)^{n}t^n + c_{n-1}t^{n-1} + \cdots + c_1t + c_0$ is the characteristic polynomial of $A$.

Notice that $\operatorname{adj}(A - tI)$ is also a poylnomial in $t$ of degree $\le n-1$, so we can pick matrices $B_0, \ldots, B_{n-1}$ such that $$\operatorname{adj}(A - tI) = \sum_{i=0}^{n-1}t^iB_i$$

Now we have \begin{align} p(t)I &= (A - tI)\operatorname{adj}(A - tI) \\ &= (A - tI)\sum_{i=0}^{n-1}t^iB_i \\ &= \sum_{i=0}^{n-1}t^i AB_i - \sum_{i=0}^{n-1}t^{i+1}B_i\\ &= -t^nB_{n-1} + \sum_{i=1}^{n-1}t^i(AB_i - B_{i-1}) + AB_0 \end{align}

Comparing powers with $p_A(t)I = (-1)^{n}t^nI + c_{n-1}I + \cdots + c_1tI + c_0I$ gives

$$B_{n-1} = (-1)^{n+1} I, \qquad AB_{i} - B_{i-1} = c_iI \text{ for } 1 \le i \le n-1, \qquad AB_0 = c_0I$$

Now we can inductively express $B_i$ as poylnomials in $A$:

$$B_{n-1} = (-1)^{n+1} I$$ $$B_{n-2} = AB_{n-1} - c_{n-1}I = (-1)^{n+1}A - c_{n-1}I$$ $$B_{n-3} = AB_{n-2} - c_{n-2}I = (-1)^{n+1}A^2 - c_{n-1}A - c_{n-2}I$$ $$\vdots$$ $$B_0 = AB_1 - c_1I = (-1)^{n+1}A^{n-1} - c_{n-1}A^{n-2} - \cdots - c_{2}A - c_1I $$

Therefore

$$\operatorname{adj}(A) = \sum_{i=0}^{n-1}t^iB_i\Bigg|_{t = 0} = B_0 = -\Big[(-1)^{n}A^{n-1} + c_{n-1}A^{n-2} + \cdots + c_{2}A + c_1I\Big]$$

Answer 2

Yes, the adjucate matrix can always be expressed as a polynomial in $A$: Consider the characteristic polynomial of $A$, i.e. $$p(A)=A^n+c_{n-1}A^{n-1}+...+c_1A+c_0=0$$ You know that $c_0=(-1)^ndet(A)$. So you can write $$det(A)=(-1)^{n+1}(A^{n-1}+c_{n-1}A^{n-2}+...+c_1)A\label{x}$$

If $det(A)\neq0$ this already implies $$A^*=(-1)^{n+1}(A^{n-1}+c_{n-1}A^{n-2}+...+c_1)$$ which is a polynomial in $A$. For $det(A)=0$, you can argue from continuity. I.e.

• The expression we derived for $A^*$ is continuous in $A$
• You know that the true $A^*$ is continuous in $A$ (because it can be expressed using co-factors)
• the set of non-invertible matrices is a measure-zero subset of all matrices

Answer 3

UPDATED TO REMOVE UNNECESSARY TOPOLOGICAL ARGUMENTS

Let $A\in\mathbb{M}_n(K)$ and let $p_A$ be its characteristic polynomial. We have $p_A(X)=Xq(X)+(-1)^n\det(A)$ for some polynomial $q\in K[X]$, so that $$(-1)^{n-1}\det(A)I_n=Aq(A).$$

Multiply by $\text{adj}(A)$ on the left to get

$$(-1)^{n-1}\det(A)\text{adj}(A)=\det(A)q(A).$$

Now consider $\det(A), q(A), \text{adj}(A)$ as polynomials in the entries of $A$, in $K[X_1,\ldots,X_{n^2}]$. Since $\det$ is a nonzero element in an integral domain, we can cancel it and get $$\text{adj}(A)=(-1)^{n-1}q(A).$$

Therefore $$\text{adj}(A)=(-1)^{n-1}\frac{p_A(X)-(-1)^n\det(A)}X\left\rvert_{X=A}\right.$$ is the formula we are seeking.